Do circuits in a changing magnetic field have a well-defined "internal electrical" potential?

Question

I am a pure mathematician and not a physicist, and so a couple of disclaimers/apologies may be in order:

Firstly, the question that I want to ask is of a sufficiently subtle nature that I need to try to be very precise about it, and the only way I can think to do that is to formulate it in a way that makes reference to advanced pure mathematics concepts. I hope this will not make the question too difficult to easily follow, and I apologise if it does.

Secondly - because of my lack of knowledge of the relevant physics - it is very possible that my question will turn out to not make physical sense (i.e. it may be based on incorrect presuppositions); if so, I apologise for that, and would be grateful for a clarification of where I've gone wrong.

---

Suppose I have a fixed loop of wire [which, by way of approximation, I'll assume to be infinitely thin] lying in the $x$-$y$ plane, such that the set of points along the wire is precisely the image of $\mathbb{R}$ under a smooth function $$ \gamma \colon \mathbb{R} \to \{(x\textrm{ metres},y\textrm{ metres},0\textrm{ metres}) : (x,y) \in \mathbb{R}^2\} $$ with the properties that

• $\gamma$ is $1$-periodic (i.e. $\gamma(\tau+1)=\gamma(\tau)$ for all $\tau \in \mathbb{R}$),
• $\gamma$ has no self-intersections on $[0,1)$ (i.e. $\gamma(\tau+\sigma) \neq \gamma(\tau)$ for all $\tau \in \mathbb{R}$ and $\sigma \in (0,1)$),
• the derivative $\gamma'$ of $\gamma$ has constant magnitude $|\gamma'(\tau)|=\ell$. [So $\ell$ is the length of the wire.]

I'm not assuming that the wire is made of a uniform material, nor that it has zero resistance all the way along its length.

Assume, by way of approximation, that there is zero inductance and zero elastance in the wire. (Is this a possible approximation?) Let $R>0$ be the total resistance of the wire. Assume that the empty space around the wire has infinite resistance.

Imagine that I'm holding a magnet, and I move my magnet along the $z$-axis so as to induce a current in the wire; and imagine that on some time-interval $[t_1,t_2]$, the time-derivative of the magnetic flux is constant, leading to a constant current through the wire. Let $I$ be the value of this constant current, measured as positive in the direction of increasing $\tau$.

Fix a time $t \in (t_1,t_2)$. Suppose (by way of approximation) that the magnitude $u$ of the drift velocity of free electrons at time $t$ is constant along the wire. (By "free electrons", I mean the electrons in the wire whose flow constitutes the current $I$, in contrast to the ions in the wire that stay fixed.) Let $$ Q = -\frac{|I|\ell}{u}. $$ (In other words, $Q$ represents the amount of free charge in the wire at time $t$.) Let $$ \mathbf{M} \colon \mathcal{B}([0,1)) \to \{(x\textrm{ newtons},y\textrm{ newtons},z\textrm{ newtons}) \, : \, (x,y,z) \in \mathbb{R}^3 \} $$ be the vector-valued measure on $[0,1)$ such that for any measurable set $A \in \mathcal{B}([0,1))$, $\mathbf{M}(A)$ is the total force exerted at time $t$ by my magnet on the free electrons occupying $\gamma(A)$.

[EDIT: I should clarify what I mean here, because this is one of the points where I suspect that my presuppositions might be false: In Newtonian mechanics, every "force" is exerted by one particle of matter on another particle of matter. For example, if at some time a particle $P$ of mass $m$ lies at a point in a gravitational field where the gravitational field strength is $\mathbf{g}$, then $m\mathbf{g}$ is equal to the vector sum of the gravitational forces exerted on $P$ by all the other particles in the system; and so for any specific subset $S$ of the set of all the other particles in the system, I can define the component of $m\mathbf{g}$ corresponding to the vector sum of the gravitational forces exerted on $P$ by the particles in $S$. I'm wanting to treat electromagnetic fields in a similar manner: If $S$ is the set of the all particles in the magnet that I'm holding, and $T$ is the set of all free electrons occupying $\gamma(A)$ at time $t$, then $\mathbf{M}(A) = \sum_{P_1 \in S} \sum_{P_2 \in T} \mathbf{f}_{P_1,P_2}$ where $\mathbf{f}_{P_1,P_2}$ is the electromagnetic force exerted by $P_1$ on $P_2$ at time $t$. So $\mathbf{M}(A)$ is not the total Lorentz force that $T$ is subjected to by virtue of the electromagnetic field that it lies in, but rather is only a "component" of that Lorentz force. Is all of this valid and well-defined, at least by way of approximation?]

My question: is it the case that $$ QIR = \int_{[0,1)} \gamma'(\tau) \boldsymbol{\cdot} \mathbf{M}(\mathrm{d}\tau) \; ? $$

(If any further [reasonably realistic] idealised approximations are needed to make the equality true, that's fine.)

---

Motivation.

The presence of a changing magnetic flux means that electrical potential is not well-defined, i.e. Kirchhoff's Voltage Law fails: letting $\mathbf{E}$ be the electric field at time $t$, we have $$ \int_0^1 \gamma'(\tau) \boldsymbol{\cdot} \mathbf{E}(\gamma(\tau)) \, \mathrm{d}\tau \neq 0 $$ even though $\gamma$ is the same at the two endpoints of the integral ($\gamma(0)=\gamma(1)$).

However: Let $$ r \colon \mathcal{B}([0,1)) \to \{x\textrm{ ohms} : x \geq 0\} $$ be the measure on $[0,1)$ such that for any measurable set $A \in \mathcal{B}([0,1))$, $r(A)$ is the resistance of the stretch of wire occupying $\gamma(A)$. (So $R=r([0,1))$.) If the answer to my question is 'yes', then at time $t$ one can define a scalar 'potential' $V$ along the wire by $$ V(\gamma(\tau)) = \frac{1}{Q}\int_{[0,\tau]} \gamma'(\sigma) \boldsymbol{\cdot} \mathbf{M}(\mathrm{d}\sigma) \, - \, Ir([0,\tau]) \quad \text{for all } \tau \in [0,1). $$ This 'potential' can be thought of as representing energy stored in the electromagnetic interactions among the free electrons and fixed ions within the wire; in other words, in loose terms, a positive answer to my question would mean that, from an energy perspective, one can decouple the electromagnetic interactions within the wire from the forces between charge in the wire and the external magnet. (At least, that's how I imagine what the implications of a positive answer to my question would be - but maybe I'm wrong.) In particular, this could provide a notion of 'potential difference' along the lines of how ElectroBOOM attempts to conceive of 'voltage' in https://youtu.be/Q9LuVBfwvzA?si=iskzAel6hIWQzPB1 (at least for my constant-current scenario - I don't know how generalisable it would be).

EDIT: In case this helps to clarify the nature of my question: this answer to a question on Electrical Engineering Stack Exchange seems to claim that in a circuit consisting simply of resistors joined by ideal wire, subject to a changing magnetic flux, there is a "partial voltage" along the stretches of ideal wire, which is equal to the negative of the path integral of the "Coulombian electric field" that is "generated by the charge that accumulates at the surface of the wire and at the interfaces with the resistive material", and which exactly negates the path integral of an "induced" electric-field contribution $E_{\mathit{ind}}$ along the same stretch of ideal wire, so that the path integral of the "total" electric field is zero along the stretch of ideal wire. My question is probably equivalent to asking: Are these claims correct, and if so, is this "Coulombian electric field generated by ..." component of the electric field a conservative field? [Well, I guess that the poster's very choice of the word "Coulombian" probably implies that the poster believes this component to be a linear superposition of inverse-square-law radially symmetric vector fields, and hence conservative.]

Answer 1

Yes, electric potential can be defined in all situations, but in a much simpler way than what you suggest.

Electric potential (let us denote it $V$ in general) is not unique; the only definitoric constraint on it is that it has to obey the equation

$$ \mathbf E = - \nabla V -\partial_t \mathbf A \tag{*}. $$

for some vector field $\mathbf A(\mathbf x, t)$, where $\mathbf E$ is the electric field in question (may be total, or partial field for which we define the potential).

There is infinity of solutions $V,\mathbf A$ in any given situation. To get a unique potential $V$, we have to introduce further constraints on $V$.

The easiest way to satisfy (*) in general (and the most used in practice, even in AC circuits) is that $V$ is given by the Coulomb integral $$ V(\mathbf x,t) = \int_{space} \frac{K\rho(\mathbf x')}{|\mathbf x-\mathbf x'|}\,d^3\mathbf x', $$ just like in electrostatics. Sometimes this is called the Coulomb electric potential, or electrostatic potential, or electric potential in the Coulomb gauge.

KVL holds in all circuits, including those with variable magnetic flux. The reason is that "voltage" in KVL is not the IEC voltage (line integral of total electric field for some path), but simply a difference of electric potential in the Coulomb gauge between two points, and sum of such differences for a closed path has to be zero, irrespective of the fact this potential does not describe the non-conservative part of total electric field.

The original Kirchhoff second circuital law is that sum of emfs in a closed path equals sum of terms $R_kI_k$ in the path. If there is only one kind of EMF in the circuit, induced EMF, we have

$$ \oint \mathbf E_i\cdot d\mathbf s = RI. $$ This is similar to the integral you ask about, except $\mathbf E_i$ is total induced electric field, not just induced electric field due to the magnet. But if self-induction can be neglected, induced field due to the circuit can be neglected, and we can write this approximative equation:

$$ \oint \mathbf E_{i,\text{due to magnet}}\cdot d\mathbf s = RI. $$

This is the same as your integral, except $\mathbf E_{i,\text{due to magnet}}$ is electric force per unit charge. If we used the actual force and integrated it, then a factor $Q$ (total mobile charge) would appear in front of $RI$, but this is not customary.

The modern KVL is different from the original Kirchhoff second circuital law, in that it uses "voltages" instead of "emfs". The difference is physically significant. Voltage in KVL is a function of two points in the circuit, and gives difference of electric potential in the Coulomb gauge; these sum up to zero in a closed path. Emfs are traditionally for the whole closed path, but can be defined for any non-closed path segments; anyway, when we sum them up for a closed path, we do not always get zero.

Despite the fact these two rules have different messages, they are both correct and lead to the same equations for circuits where external EMFs are negligible (e.g. due to external variable magnetic flux), because we can express all potential differences in terms of the other quantities at play.

If external EMF is not negligible, like in your example with a moving magnet, then KVL, while it is correct, is not applicable (useful), because we can't express its voltages (=potential differences) as functions of $Q,I,\dot{I}$ and emfs at play; their exact values remain unknown, because they depend on details of the geometry of the circuit and charge distribution in the source of EMF which is unspecified and not really that important. Then the original formulation (second circuital law with emfs) is superior, because it does not depend on those unspecified details, and gives the correct circuit equation.

Answer 2

As @naturallyInconsistent commented, in the case of a changing magnetic field the ‘electric’ field is ${\bf E} = -\dot{\bf A}$. The potential that the OP is looking for is the vector potential $\bf A$.

In the standard, gauge invariant theory the scalar and vector potentials are determined only up to a gauge transformation. However, only the Lorenz gauge guarantees that the potentials are causal, so I strongly advise against using any other gauge.